From exam preperation. I consider the operator $T f(x) := f(x − 1), x ∈ \mathbb{R}$. First on the space of all function $f:\mathbb{R}\rightarrow \mathbb{C}$. There I found that any number other than $0$ is an eigenvalue. Now it is claimed in the exercise that $T$ has no eigenvalues when $T : L^2 (\mathbb{R}) → L^ 2 (\mathbb{R})$.
Why is that? How does it change the argument, that the function is in $L^ 2(\mathbb{R})$?
On
Let $f$ be an eigenfunction to an eigenvalue $\lambda$. Since $f\in L^2$, we have $$ \|f\|_{L^2}^2 = \sum_{k\in \mathbb Z}\int_{(k,k+1)} f^2 dx= \lim_{n\to\infty}\sum_{k\in \mathbb Z:-n\le k<n}\int_{(k,k+1)} f^2 dx. $$ Due to the eigenvalue property, $$ \int_{(n,n+1)} f^2 dx = \lambda^n \int_{(0,1)} f^2dx $$ for all $n$, which implies $$ \|f\|_{L^2}^2 = \int_{(0,1)} f^2dx \cdot \lim_{n\to\infty}\sum_{k\in \mathbb Z:-n\le k<n}\lambda^k. $$ The left-hand side is a finite number, so the right-hand side has to be finite, which implies $\int_{(0,1)} f^2dx =0$ or $\lambda=0$. Both imply $f=0$, which is a contradiction $f$ being an eigenfunction.
Because constant functions, except $0$, are not elements of $L^2(\mathbb{R})$ as they have an infinite integral.